converting into date formats

[julian parodi]

Hi, how do I convert these variables into actual dates? thank you!

[Nick Cox]

year must be a numeric variable if you successfully assigned it a daily date display format. It would seem to follow that

1. There is no need to create another daily date variable from it.

2. Any attempt to do that using date() is illegal as the latter expects a string argument.

3. The name year appears to be a poor choice.

However, it is also possible that this answer misses your point. You don’t give a convincing data example (do please read and act on FAQ Advice #12) and the request for “actual dates” is not at all clear to me.

[julian parodi]

Hi, sorry, the variables are currently %td and I was wondering how to just change them into YYYY, to take the 31dec away! TY

[Rich Goldstein]

if you look at the help file (-help datetime-) in the section on displaying, you will see that %ty is meant for that purpose; use the -format- command to change the display format

[Nick Cox]

As Rich Goldstein says, you need to read the help:

help datetime

Changing the display format is not sufficient here, or even necessary. Let's take today's date:

. di %td  mdy(11, 26, 2019)
26nov2019

So far, so good. But change the display format to year and you are saying that the date is really a year: The display result will puzzle a little because Stata has no expectation of 4 digit years, but in the case of today's date if you change the format you are telling Stata that the year is really 21879.

. di %ty  mdy(11, 26, 2019)
2.2e+04

. di %5.0f  mdy(11, 26, 2019)
21879

But it's not. You need a conversion function and it can just be year():

. di %td  mdy(11, 26, 2019)
26nov2019

. di %ty  mdy(11, 26, 2019)
2.2e+04

. di %5.0f  mdy(11, 26, 2019)
21879

. di year(mdy(11, 26, 2019))
2019

So, push your variable through year().

[William Lisowski]

Let me add to Nick's response a note that if you read help datetime and the pages linked from it, you will learn that it is possible to change the format of a daily date like 31Dec2017 to only display the year portion. But that doesn't solve your analytical problems, because the underlying value is still the same, you have just changed how Stata displays the number.

You (apparently) have yearly data measured on the last day of each year. If you have a variable, say X, in an observation with a daily date of 31Dec2017, and you specify L.X to obtain the lagged value of X, Stata will seek the observation with a date of 30December2017, because it recognizes your dates as daily dates and looks for the previous day.

So you need a yearly date variable as Nick describes to ensure that calculations that depend on relative times, like leads and lags, will do what you want.

[julian parodi]

Thank you Rich, Nick and William!
format %ty year
format %5.0f year
replace year=year/365
worked

[William Lisowski]

Consider the following example

* Example generated by -dataex-. To install: ssc install dataex
clear
input float date
20453
20819
21184
end
format %td date

generate year = yofd(date)
generate badyear = date/365
list, clean noobs

. list, clean noobs

         date   year    badyear  
    31dec2015   2015   56.03562  
    31dec2016   2016   57.03836  
    31dec2017   2017   58.03836

Following the techniques in the documentation cited in post #6 is advisable if you want to create a year that is meaningful to Stata. Note that the year() function Nick used in post #5 is another name for the yofd() function in the example above and in the documentation.

[Nick Cox]

As William Lisowski implies #7 really was not recommended and is a thoroughly bad idea. Let readers beware. We decided against downvotes here when converting to a web forum, but occasionally they would be useful.